Given the matrix \(A = \begin{pmatrix} -2 & 3 & 1 \\ p & 2 & 1 \\ 1 & 4 & 2 \end{pmatrix}\) the determinant is set to \(-5\):

\(\text{det}(A) = -2 \begin{vmatrix} 2 & 1 \\ 4 & 2 \end{vmatrix} - 3 \begin{vmatrix} p & 1 \\ 1 & 2 \end{vmatrix} + 1 \begin{vmatrix} p & 2 \\ 1 & 4 \end{vmatrix}\)

Calculating the \(2 \times 2\) determinants:

1. \( \begin{vmatrix} 2 & 1 \\ 4 & 2 \end{vmatrix} = 0 \)
2. \( \begin{vmatrix} p & 1 \\ 1 & 2 \end{vmatrix} = 2p - 1 \)
3. \( \begin{vmatrix} p & 2 \\ 1 & 4 \end{vmatrix} = 4p - 2 \)

So, \(\text{det}(A) = -3(2p - 1) + (4p - 2) = -6p + 3 + 4p - 2 = -2p + 1\)

Setting this equal to \(-5\): \(-2p + 1 = -5\)

Solving for \( p \): \(-2p = -6 \implies p = 3\)

Thus, the value of \( p \) is 3